# How do you find the derivative of y= cos^3 w + cos(w^3)?

##### 1 Answer
Mar 5, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 \sin w {\cos}^{2} w - 3 {w}^{2} \sin \left({w}^{3}\right)$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left[a + b + c\right] = \frac{d}{\mathrm{dx}} \left[a\right] + \frac{d}{\mathrm{dx}} \left[b\right] + \frac{d}{\mathrm{dx}} \left[c\right]$

so

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[{\cos}^{3} w\right] + \frac{d}{\mathrm{dx}} \left[\cos \left({w}^{3}\right)\right]$

Now we've split it up, we can tackle each term separately.

The product rule states that

$\frac{d}{\mathrm{dx}} a b = b \frac{d}{\mathrm{dx}} \left[a\right] + a \frac{d}{\mathrm{dx}} \left[b\right]$

so

$\frac{d}{\mathrm{dx}} {\cos}^{3} w = \cos w \frac{d}{\mathrm{dx}} \left[{\cos}^{2} w\right] + {\cos}^{2} w \frac{d}{\mathrm{dx}} \cos w$

$\frac{d}{\mathrm{dx}} {\cos}^{2} w = \cos w \frac{d}{\mathrm{dx}} \cos x + \cos w \frac{d}{\mathrm{dx}} \cos w$

$= - 2 \sin w \cos w$

therefore,

$\frac{d}{\mathrm{dx}} {\cos}^{3} w = \cos w \cdot - 2 \sin w \cos w + {\cos}^{2} w \cdot - \sin w$

$= - 3 \sin w {\cos}^{2} w$

Now we can begin to look at the second term, for which we need the chain rule:

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = g ' \left(x\right) f ' \left(g \left(x\right)\right)$

so

$\frac{d}{\mathrm{dx}} \cos \left({w}^{3}\right) = 3 {w}^{2} \cdot - \sin \left({w}^{3}\right) = - 3 {w}^{2} \sin \left({w}^{3}\right)$

Now we can put the whole thing back together,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 \sin w {\cos}^{2} w - 3 {w}^{2} \sin \left({w}^{3}\right)$